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B
51C7+30C7
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C
50C7−30C7
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D
50C6−30C6
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Solution
The correct option is A51C7−30C7 We have, 20∑r=050−rC6 =50C6+49C6+48C6+⋯+30C6 =−30C7+30C7+30C6+31C6+⋯+49C6+50C6 =−30C7+31C7+31C6+⋯+49C6+50C6 =−30C7+32C7+32C6+⋯+49C6+50C6
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